So far, no one has voiced objections, so should I go ahead and check
this in?
btw, thanks Mike, what about this one:
>>> (np.char.lstrip(c, ' ') == np.char.lstrip(c, '')).all()
... # XXX: is this a regression? this line now returns False -pi
... # np.char.lstrip(c,'') does not modify
On Sun, Dec 13, 2009 at 00:37, David Cournapeau wrote:
> On Sun, Dec 13, 2009 at 4:27 AM, Chris wrote:
>> Here is a log form a build of svn rev 7996 with no LDFLAGS specified, as
>> recommended by Robert. The result is the same, however.
>>
>> http://files.me.com/fonnesbeck/y7e9v2
>
> I don't see
On Sun, Dec 13, 2009 at 4:27 AM, Chris wrote:
> Here is a log form a build of svn rev 7996 with no LDFLAGS specified, as
> recommended by Robert. The result is the same, however.
>
> http://files.me.com/fonnesbeck/y7e9v2
I don't see any build error on this log ?
David
___
Have you considered creating a TimeSeries for each data series, and
then putting them all together in a dict, keyed by symbol?
One disadvantage of one big monster numpy array for all the series is
that not all series may have a full set of 1800 data points. So the
array isn't really nicely
On Sat, Dec 12, 2009 at 8:08 PM, THOMAS BROWNE wrote:
> Hello all,
>
> Quite new to numpy / timeseries module, please forgive the elementary
> question.
>
> I wish to do quite to do a bunch of multivariate analysis on 1000 different
> financial markets series, each holding about 1800 data points
Hello all,
Quite new to numpy / timeseries module, please forgive the elementary question.
I wish to do quite to do a bunch of multivariate analysis on 1000 different
financial markets series, each holding about 1800 data points (5 years of daily
data).
What's the best way to put this into a
Here is a log form a build of svn rev 7996 with no LDFLAGS specified, as
recommended by Robert. The result is the same, however.
http://files.me.com/fonnesbeck/y7e9v2
cf
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2009/12/12 T J :
> Hi,
>
> Suppose I have an array of shape: (n, k, k). In this case, I have n
> k-by-k matrices. My goal is to compute the product of a (potentially
> large) user-specified selection (with replacement) of these matrices.
> For example,
>
> x = [0,1,2,1,3,3,2,1,3,2,1,5,3,2,3,5,
Le samedi 12 décembre 2009 à 01:55 -0800, T J a écrit :
> Is there a better way?
You may have a look at http://scipy.org/Cookbook/MultiDot
Several alternatives are proposed.
Cheers,
--
Denis
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On Sat, Dec 12, 2009 at 4:55 PM, T J wrote:
> Hi,
>
> Suppose I have an array of shape: (n, k, k). In this case, I have n
> k-by-k matrices. My goal is to compute the product of a (potentially
> large) user-specified selection (with replacement) of these matrices.
> For example,
>
> x = [0,1,
Hi,
Suppose I have an array of shape: (n, k, k). In this case, I have n
k-by-k matrices. My goal is to compute the product of a (potentially
large) user-specified selection (with replacement) of these matrices.
For example,
x = [0,1,2,1,3,3,2,1,3,2,1,5,3,2,3,5,2,5,3,2,1,3,5,6]
says that I
2009/12/12 Ernest Adrogué :
> Hi,
>
> Suppose I have a flat array, and I want to know the
> indices corresponding to values contained in a list
> of arbitrary lenght.
>
> Intuitively I would have done:
>
> a = np.array([1,2,3,4])
> np.nonzero(a in (0,2,4))
>
> However the "in" operator doesn't work
On Sat, Dec 12, 2009 at 5:59 AM, Jasper van de Gronde
wrote:
> Francesc Alted wrote:
>> ...
>> Yeah, I think taking slices here is taking quite a lot of time:
>>
>> In [58]: timeit E + Xi2[P/2,:]
>> 10 loops, best of 3: 3.95 µs per loop
>>
>> In [59]: timeit E + Xi2[P/2]
>> 10 loops, best
np.setmember1d(a,b)
does the same as your
reduce(np.logical_or, [a == i for i in b])
but it's actually slower on my machine!
Gary R.
Ernest Adrogué wrote:
> Hi,
>
> Suppose I have a flat array, and I want to know the
> indices corresponding to values contained in a list
> of arbitrary lenght.
Hi,
Suppose I have a flat array, and I want to know the
indices corresponding to values contained in a list
of arbitrary lenght.
Intuitively I would have done:
a = np.array([1,2,3,4])
np.nonzero(a in (0,2,4))
However the "in" operator doesn't work element-wise,
instead it compares the whole arr
Francesc Alted wrote:
> ...
> Yeah, I think taking slices here is taking quite a lot of time:
>
> In [58]: timeit E + Xi2[P/2,:]
> 10 loops, best of 3: 3.95 µs per loop
>
> In [59]: timeit E + Xi2[P/2]
> 10 loops, best of 3: 2.17 µs per loop
>
> don't know why the additional ',:' in the
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